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I found this approximation of which an earlier version I posted in the chat room:

$$7 \pi -\text{Log}\left[\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi \right] = 14.13472514154629716253329494571302508888...$$

The first non trivial zeta zero: $$14.13472514173469379045725198356247027078$$

Can you improve on the formula above?

Edit 2.9.2012

Based on the comments below I would like to explain how I reasoned:

Any Taylor series evaluated at $x=1$ is convergent for variants of it when multiplied element wise with rows in this matrix:

$$\begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$

Many Taylor series have the second row as part of its coefficients. That is: $$(1,-1,1,-1,1,-1,1,-1,1,-1,...)$$

Such Taylor series are for example $\log 2$, $\sqrt 2$, $\cos 1$, $\sin 1$. The reason for the convergence of such series and divisibility defined variants of thereof, seems to be that in the matrix above, a period sums to zero.

The simplest Dirichlet series that sums to zero and is not a an element wise multiplication of two other Dirichlet series, is the first row:

$$\frac{0}{1}+\frac{0}{2}+\frac{0}{3}+\frac{0}{4}+\frac{0}{5}+... = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

This suggests that one should try to find an expression for a such sequence.

The definition of a number raised to a complex number is:

$$n^{(a+ib)} = n^{a}(\cos (b \log (n))+i\sin (b \log (n)))$$

and the Riemann zeta function is:


where $s$ is a complex number.

Here I then made a mistake. I started studying the equation: $$\cos (\log (n)) = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ in order to get something similar to the Dirichlet series with numerators equal to the all zeros sequence in expression $(1)$ above. But if I understand correctly this would be the same as seeking the undefined sequence:


After that I just guessed that by combining values from the solutions to equation $(2)$ one could possibly find an expression for the zeta zeros.

Edit 23.12.2012: For what it is worth. Here is how the actual calculation went:

The first Riemann zeta zero is:

$$\Im(\rho _1)$$ $$=14.1347251417346937904572519836$$

A number close to the first Riemann zeta zero is:

$$\frac{9 \pi }{2}$$ $$=14.1371669411540695730818952248$$

That number can be split up into:

$$\frac{9 \pi }{2} = 7 \pi -\log \left(e^{\frac{5 \pi }{2}}\right)$$

To see what is missing within the logarithm I added an $x$ and solved the equation:

$$\text{Solve}\left[N\left[7 \pi -\log \left(x+e^{\frac{5 \pi }{2}}\right),30\right]=N[\Im(\rho _1),30],x\right]$$

This gives the solution:

$$\{\{x\to 6.297688980465813720589098\}\}$$

which is close to:

$$2\pi = 6.28318530717958647692528676656...$$

Substituting $x$ with $2\pi$:

$$7 \pi -\log \left(e^{\frac{5 \pi }{2}}+2 \pi \right)$$

which is closer:


Some small number seems to be missing, the second harmonic number could be it:

$$7 \pi -\log \left(e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

which again is closer:


Multiplying the added number with $\frac{3}{2}$

$$7 \pi -\log \left(\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

closer still:


continuing the trick with similar numbers:

$$7 \pi -\log \left(\frac{5}{2} e^{-\frac{1}{2} (5 \pi )}+\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$



and once more:

$$7 \pi -\log \left(\frac{7}{2} e^{-\frac{1}{2} (7 \pi )}+\frac{5}{2} e^{-\frac{1}{2} (5 \pi )}+\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

it works:


but then I can't get further.

Edit: 5.11.2013:

$$\frac{\sqrt{\frac{\Im(\rho _1)}{\pi }+\frac{1}{2}}}{\sqrt{5}}=0.999922272089659461895288929782$$

$$\frac{\Im(\rho _1)}{\pi }+\frac{1}{2}=4.99922275110473484848654142318$$

$\rho _1$ = first riemann zeta zero = 14.134725141734693790457...

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You might want to add some explanations about this exact-approximate formula (the one with $\pi$ and exponentials). Is it composed of the first terms of an explicit series? – Did Sep 2 '12 at 17:01
Interesting, but $2171956\over153661$ is closer. I mean, unless the pattern lurking in your exression leads to a convergent sequence (which seems is not the case), this looks quite fruitless to me. – Hagen von Eitzen Sep 2 '12 at 17:02
Agreed, unless you can come up with a "reason," this could easily just be a coincidence. – Thomas Andrews Sep 2 '12 at 17:06
I will try to explain by editing the question above, with a Edit section below the original question. – Mats Granvik Sep 2 '12 at 17:11
I fail to see what the Edit section explains. – Did Sep 2 '12 at 18:15

Can you improve on the formula above? -- Yes: $$ \text{Log}\left[\frac{2}{3} e^{-5 \pi /2}+\frac{e^{7 \pi }}{\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi }\right] = 14.1347251417343... $$

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